Problem 47 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A vector of magnitude 2 along a bisector of the angle between the two vectors \(2 i-2 j+k\) and \(i+2 j-2 k\) is (A) \(\frac{2}{\sqrt{10}}(3 i-k)\) (B) \(\frac{1}{\sqrt{26}}(i-4 j+3 k)\) (C) \(\frac{2}{\sqrt{26}}(i-4 j+3 k)\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (A) \(\frac{2}{\sqrt{10}}(3i - k)\).

Step 1: Determine Unit Vectors

Let's first find the unit vectors of the given vectors. The first is \(\vec{a} = 2i - 2j + k\), and the second is \(\vec{b} = i + 2j - 2k\). The magnitude of \(\vec{a}\) is calculated using the formula \(\|\vec{a}\| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3\), making the unit vector \(\hat{a} = \frac{1}{3}(2i - 2j + k)\). For \(\vec{b}\), its magnitude is \(\|\vec{b}\| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3\), leading to the unit vector \(\hat{b} = \frac{1}{3}(i + 2j - 2k)\).

Step 2: Find the Bisector Vector

The bisector of the angle between two vectors \(\vec{a}\) and \(\vec{b}\) is given by the sum of their unit vectors. Thus, the bisector \(\vec{c}\) is \(\hat{a} + \hat{b} = \frac{1}{3}(2i - 2j + k) + \frac{1}{3}(i + 2j - 2k)\). Simplifying, \(\vec{c} = \frac{1}{3}(3i + 0j - k) = i - \frac{1}{3}k\).

Step 3: Calculate the Magnitude of Bisector

Next, find the magnitude of \(\vec{c}\), which is calculated as follows: \(\|\vec{c}\| = \sqrt{1^2 + 0^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{1 + \frac{1}{9}} = \sqrt{\frac{10}{9}} = \frac{\sqrt{10}}{3}\).

Step 4: Formulate the Required Vector With Given Magnitude

To find a vector that is along the bisector and has a magnitude of 2, we scale \(\vec{c}\) by \(\frac{2}{\|\vec{c}\|}\) resulting in \(\frac{2}{\frac{\sqrt{10}}{3}}(i - \frac{1}{3}k) = \frac{6}{\sqrt{10}}(i - \frac{1}{3}k)\). This simplifies to \(\frac{2}{\sqrt{10}}(3i - k)\), which matches option (A).

Key Concepts

Vector Magnitude

In vector algebra, the magnitude of a vector—also known as its length or norm—tells us how long the vector is. It's like measuring the distance a vector covers from the origin to its endpoint in the vector space.

To calculate the magnitude of a vector, you’ll use the formula:

• For vector \( \vec{a} = ai + bj + ck \), the magnitude is \( \|\vec{a}\| = \sqrt{a^2 + b^2 + c^2} \).

This formula finds the 'straight line' distance from the start to the end of the vector, similar to finding the length of a side in a 3-dimensional space.
For example, if you have a vector \( 2i - 2j + k \), you can determine its magnitude by substituting into the formula: \( \|\vec{a}\| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3 \).

This value represents the distance our vector covers. Knowing the magnitude is essential for tasks like normalizing a vector, which we'll cover in more detail in the section about unit vectors.

Angle Bisector

The angle bisector is an important concept when you are analyzing the relationship between two vectors. Specifically, an angle bisector in vector terms is a vector that divides the angle between two vectors into two equal parts.

To find the bisector vector of two vectors \( \vec{a} \) and \( \vec{b} \), you add their unit vectors together. Here’s the process:

• First, determine the unit vectors for \( \vec{a} \) and \( \vec{b} \).
• Sum the unit vectors to find the bisector.

For instance, if we have \( \vec{a} = 2i - 2j + k \) and \( \vec{b} = i + 2j - 2k \), calculate their unit vectors first, then add them:

• \( \hat{a} = \frac{1}{3}(2i - 2j + k) \)
• \( \hat{b} = \frac{1}{3}(i + 2j - 2k) \)
• The bisector \( \vec{c} = \hat{a} + \hat{b} = i - \frac{1}{3}k \)

This bisector vector represents a direction at the midpoint between the input vectors, perfect for scenarios where symmetrical division is required.

Unit Vectors

Unit vectors are immensely useful in vector algebra as they provide a way to describe direction without considering magnitude. They are vectors with a length of exactly one.

To convert any vector into a unit vector, you simply divide the vector by its magnitude. The resulting vector maintains the same direction but is scaled to a length of one, making it a 'standard' reference direction.
Here's how you find a unit vector:

• For a vector \( \vec{a} = ai + bj + ck \), the unit vector \( \hat{a} \) is \( \frac{1}{\|\vec{a}\|}(ai + bj + ck) \).

In practice, if we take the vector \( 2i - 2j + k \), its magnitude is 3. The unit vector then becomes:

• \( \hat{a} = \frac{1}{3}(2i - 2j + k) \)

Unit vectors are critical when defining directions, as they can be used to scale the vector up or down to any desired length simply by multiplying by the needed magnitude. They map perfectly to using the angle bisector concept, ensuring that vectors can be easily combined.